Finding derivative of $f(x)$ where $f(xy) = f(x) + f(y)$ - without change of variable












1














Let $f(x)$ be a function $(0,infty) to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$



It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = frac{1}{x} f'(1)$.



Solution:



I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:



$f'(1) = lim_{hto 0} frac{f(1+h) - f(1)}{h} = lim_{hto 0}frac{f(1+h)}{h} $



so



$f'(x) = lim_{hto 0} frac{f(x+h) - f(x)}{h} = lim_{hto 0} frac{f(frac{x+h}{x})}{h} = lim_{hto 0} frac{f(1 + h/x)}{h}$.



I know I can solve it with a simple change of variables $h to 0 ~~~rightarrow~~~~ h/x to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?






limits derivatives logarithms change-of-variable







share|cite|improve this question


















  • 1




    @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
    – Umberto P.
    Nov 26 at 18:44












  • @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
    – amir na
    Nov 26 at 18:49










  • OK, I understand now.
    – kimchi lover
    Nov 26 at 19:05
















1














Let $f(x)$ be a function $(0,infty) to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$



It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = frac{1}{x} f'(1)$.



Solution:



I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:



$f'(1) = lim_{hto 0} frac{f(1+h) - f(1)}{h} = lim_{hto 0}frac{f(1+h)}{h} $



so



$f'(x) = lim_{hto 0} frac{f(x+h) - f(x)}{h} = lim_{hto 0} frac{f(frac{x+h}{x})}{h} = lim_{hto 0} frac{f(1 + h/x)}{h}$.



I know I can solve it with a simple change of variables $h to 0 ~~~rightarrow~~~~ h/x to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?






limits derivatives logarithms change-of-variable







share|cite|improve this question


















  • 1




    @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
    – Umberto P.
    Nov 26 at 18:44












  • @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
    – amir na
    Nov 26 at 18:49










  • OK, I understand now.
    – kimchi lover
    Nov 26 at 19:05














1












1








1







Let $f(x)$ be a function $(0,infty) to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$



It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = frac{1}{x} f'(1)$.



Solution:



I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:



$f'(1) = lim_{hto 0} frac{f(1+h) - f(1)}{h} = lim_{hto 0}frac{f(1+h)}{h} $



so



$f'(x) = lim_{hto 0} frac{f(x+h) - f(x)}{h} = lim_{hto 0} frac{f(frac{x+h}{x})}{h} = lim_{hto 0} frac{f(1 + h/x)}{h}$.



I know I can solve it with a simple change of variables $h to 0 ~~~rightarrow~~~~ h/x to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?






limits derivatives logarithms change-of-variable







share|cite|improve this question













Let $f(x)$ be a function $(0,infty) to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$



It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = frac{1}{x} f'(1)$.



Solution:



I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:



$f'(1) = lim_{hto 0} frac{f(1+h) - f(1)}{h} = lim_{hto 0}frac{f(1+h)}{h} $



so



$f'(x) = lim_{hto 0} frac{f(x+h) - f(x)}{h} = lim_{hto 0} frac{f(frac{x+h}{x})}{h} = lim_{hto 0} frac{f(1 + h/x)}{h}$.



I know I can solve it with a simple change of variables $h to 0 ~~~rightarrow~~~~ h/x to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?






limits derivatives logarithms change-of-variable




limits derivatives logarithms change-of-variable






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 18:22









amir na

385




385








  • 1




    @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
    – Umberto P.
    Nov 26 at 18:44












  • @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
    – amir na
    Nov 26 at 18:49










  • OK, I understand now.
    – kimchi lover
    Nov 26 at 19:05














  • 1




    @kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
    – Umberto P.
    Nov 26 at 18:44












  • @kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
    – amir na
    Nov 26 at 18:49










  • OK, I understand now.
    – kimchi lover
    Nov 26 at 19:05








1




1




@kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
– Umberto P.
Nov 26 at 18:44






@kimchilover it is stated explicitly that $f$ is differentiable at $x=1$.
– Umberto P.
Nov 26 at 18:44














@kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
– amir na
Nov 26 at 18:49




@kimchilover it is said in the body of the question that it is differentiable at 1. It is not my deduction.
– amir na
Nov 26 at 18:49












OK, I understand now.
– kimchi lover
Nov 26 at 19:05




OK, I understand now.
– kimchi lover
Nov 26 at 19:05










2 Answers
2






active

oldest

votes


















2














You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
$$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
Thus
$$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
and consequently
$$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$






share|cite|improve this answer























  • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
    – amir na
    Nov 26 at 18:47










  • Please see the edited answer.
    – Umberto P.
    Nov 26 at 18:49










  • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
    – A.Γ.
    Nov 26 at 18:51












  • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
    – amir na
    Nov 26 at 18:54










  • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
    – amir na
    Nov 26 at 18:57



















1














Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
begin{align}
frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
&=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
end{align}






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014713%2ffinding-derivative-of-fx-where-fxy-fx-fy-without-change-of-var%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



    A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




    The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
    $$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
    Thus
    $$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




    Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
    and consequently
    $$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$






    share|cite|improve this answer























    • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
      – amir na
      Nov 26 at 18:47










    • Please see the edited answer.
      – Umberto P.
      Nov 26 at 18:49










    • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
      – A.Γ.
      Nov 26 at 18:51












    • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
      – amir na
      Nov 26 at 18:54










    • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
      – amir na
      Nov 26 at 18:57
















    2














    You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



    A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




    The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
    $$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
    Thus
    $$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




    Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
    and consequently
    $$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$






    share|cite|improve this answer























    • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
      – amir na
      Nov 26 at 18:47










    • Please see the edited answer.
      – Umberto P.
      Nov 26 at 18:49










    • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
      – A.Γ.
      Nov 26 at 18:51












    • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
      – amir na
      Nov 26 at 18:54










    • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
      – amir na
      Nov 26 at 18:57














    2












    2








    2






    You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



    A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




    The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
    $$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
    Thus
    $$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




    Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
    and consequently
    $$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$






    share|cite|improve this answer














    You pointed out that $f'(1) = displaystyle lim_{h to 0} frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.



    A simple substitution shows that $displaystyle lim_{h to 0} frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.




    The above statement follows from a basic $epsilon$ argument. Let $epsilon > 0$ be given. There exists $delta > 0$ with the property that
    $$0 < |h| < delta implies left| frac{f(1+h)}{h} - f'(1) right| < epsilon.$$
    Thus
    $$0 < |h| < x delta implies 0 < left| frac hx right| < delta implies left| frac{f(1+h/x)}{h/x} - f'(1) right| < epsilon$$ so that $$frac{f(1+h/x)}{h/x} to f'(1).$$




    Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$frac{f(x+h) - f(x)}{h} = frac{f(1 + h/x)}{h} = frac 1x frac{f(1 + h/x)}{h/x}$$
    and consequently
    $$lim_{h to 0} frac{f(x+h) - f(x)}{h} = frac 1x lim_{h to 0} frac{f(1 + h/x)}{h/x} = frac 1x cdot f'(1).$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 26 at 18:53

























    answered Nov 26 at 18:42









    Umberto P.

    38.5k13064




    38.5k13064












    • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
      – amir na
      Nov 26 at 18:47










    • Please see the edited answer.
      – Umberto P.
      Nov 26 at 18:49










    • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
      – A.Γ.
      Nov 26 at 18:51












    • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
      – amir na
      Nov 26 at 18:54










    • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
      – amir na
      Nov 26 at 18:57


















    • As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
      – amir na
      Nov 26 at 18:47










    • Please see the edited answer.
      – Umberto P.
      Nov 26 at 18:49










    • @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
      – A.Γ.
      Nov 26 at 18:51












    • I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
      – amir na
      Nov 26 at 18:54










    • @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
      – amir na
      Nov 26 at 18:57
















    As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
    – amir na
    Nov 26 at 18:47




    As I said in the question, I know it can be solved like this, But in this solution it is assumed that $hto0$ means $h/x to 0$. but someone told me when you don't know that the function is continuous, you can't simply substitute variables (i.e. h changed to h/x) . I want to know can it be proved just by definition of limit ($epsilon , delta$)?
    – amir na
    Nov 26 at 18:47












    Please see the edited answer.
    – Umberto P.
    Nov 26 at 18:49




    Please see the edited answer.
    – Umberto P.
    Nov 26 at 18:49












    @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
    – A.Γ.
    Nov 26 at 18:51






    @amirna When $hto 0$ and $xne 0$ then $frac{h}{x}to 0$. It is an easy exercise on $epsilon-delta$: $left|frac{h}{x}right|<epsilon$ iff $|h|<epsilon|x|=delta$.
    – A.Γ.
    Nov 26 at 18:51














    I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
    – amir na
    Nov 26 at 18:54




    I saw your answer and I know it is completely valid. However, I want to know whether this question can be solved with definition of limit or not. I mean getting $epsilon,delta$ from f'(1) (because we know it exists) and then changing them somehow for using in the $f'(x)$ definition/
    – amir na
    Nov 26 at 18:54












    @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
    – amir na
    Nov 26 at 18:57




    @A.Γ. I saw your comment now, it seems fine but I think maybe it must be $|h/x| < delta '$. because actually $epsilon$ is used in $|f(x) - L| < epsilon$ part of the definition and not in the boundary of the variable.
    – amir na
    Nov 26 at 18:57











    1














    Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
    begin{align}
    frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
    &=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
    end{align}






    share|cite|improve this answer


























      1














      Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
      begin{align}
      frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
      &=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
      end{align}






      share|cite|improve this answer
























        1












        1








        1






        Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
        begin{align}
        frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
        &=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
        end{align}






        share|cite|improve this answer












        Set $y=1$ to verify that $f(1)=0$. Now when $hto 0$
        begin{align}
        frac{f(x+h)-f(x)}{h}&=frac{f((1+h/x)x)-f(x)}{h}=frac{f(1+h/x)+f(x)-f(x)}{h}=\
        &=frac{f(1+h/x)}{h/x}cdot frac{1}{x}=frac{f(1+h/x)-f(1)}{h/x}cdotfrac{1}{x}to ?
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 26 at 18:38









        A.Γ.

        21.8k22455




        21.8k22455






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014713%2ffinding-derivative-of-fx-where-fxy-fx-fy-without-change-of-var%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more